Q138 / 155 · WBJEE 2018 · Physics
A point charge $-q$ is carried from a point $A$ to another point $B$ on the axis of a charged ring of radius $r$ carrying a charge $+q$. If the point $A$ is at a distance $\frac{4}{3} r$ from the centre of the
ring and the point $B$ is $\frac{3}{4} r$ from the centre but
on the opposite side, what is the net work that need to be done for this?
- A$-\frac{7}{5} \cdot \frac{q^{2}}{4 \pi \varepsilon_{0} r}$
- Correct answerB$-\frac{1}{5} \cdot \frac{q^{2}}{4 \pi \varepsilon_{0} I}$
- C$\frac{7}{5} \cdot \frac{q^{2}}{4 \pi \varepsilon_{0} r}$
- D$\frac{1}{5} \cdot \frac{q^{2}}{4 \pi \varepsilon_{0} r}$
Solution
A charged circular ring of radius $R$ is shown in figure.
From the figure, $d=\sqrt{x^{2}+r^{2}}$
$\therefore$ Potential due to a uniform ring of positive charge $q$ at point $P$
$$
\begin{aligned}
V &=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{d} \\
&=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{\sqrt{x^{2}+r^{2}}} \quad[\text { from Eq. }
\end{aligned}
$$
Now,
$\therefore C B=\sqrt{O C^{2}+O B^{2}}=\sqrt{r^{2}+\left(\frac{3}{4}\right)^{2} r^{2}}=\frac{5}{4} r$
Similarly, $A C=\frac{5}{3} r$
$\therefore$ Work done in bringing a point charge $-q$ from $A$
to $B$,
$\begin{aligned} W &=-q\left(V_{B}-V_{A}\right) \\ &=-q\left[\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{C B}-\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{C A}\right] \\=& \frac{-1}{4 \pi \varepsilon_{0}} \cdot q^{2}\left[\frac{1}{\frac{5}{4} r}-\frac{1}{5}\right] \\=& \frac{-1}{4 \pi \varepsilon_{0}} \cdot \frac{q^{2}}{5 r}[4-3]=-\frac{1}{5} \cdot \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q^{2}}{r} \end{aligned}$
From the figure, $d=\sqrt{x^{2}+r^{2}}$
$\therefore$ Potential due to a uniform ring of positive charge $q$ at point $P$
$$
\begin{aligned}
V &=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{d} \\
&=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{\sqrt{x^{2}+r^{2}}} \quad[\text { from Eq. }
\end{aligned}
$$
Now,
$\therefore C B=\sqrt{O C^{2}+O B^{2}}=\sqrt{r^{2}+\left(\frac{3}{4}\right)^{2} r^{2}}=\frac{5}{4} r$
Similarly, $A C=\frac{5}{3} r$
$\therefore$ Work done in bringing a point charge $-q$ from $A$
to $B$,
$\begin{aligned} W &=-q\left(V_{B}-V_{A}\right) \\ &=-q\left[\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{C B}-\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{C A}\right] \\=& \frac{-1}{4 \pi \varepsilon_{0}} \cdot q^{2}\left[\frac{1}{\frac{5}{4} r}-\frac{1}{5}\right] \\=& \frac{-1}{4 \pi \varepsilon_{0}} \cdot \frac{q^{2}}{5 r}[4-3]=-\frac{1}{5} \cdot \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q^{2}}{r} \end{aligned}$Hence, the correct answer is option: 'B'.